Follow one fictional student’s full attempt: the working, the marks it earned, and what to fix next. All 20 questions are here, including the mistakes.
The sequence begins 7, 11, 15, 19, … Find an expression for its nth term and find its 20th term.
Student working
Common difference = 4✓
nth term = 4n + 3✓
20th term = 4 × 20 + 3 = 87×
Student’s final answer
4n + 3; 87
Marking breakdown
1/1
Correct constant difference.
2/2
Correct coefficient and offset: n = 1 gives 7.
0/1
80 + 3 is 83, not 87.
Correct answer
4n + 3; 83
Step-by-step explanation +
The rule is correct, so the method marks are retained. The 20th term is 4(20) + 3 = 83. Recheck simple arithmetic even after a correct algebraic method.
A 13 m ladder leans against a vertical wall. Its foot is 5 m from the wall on level ground. Find the height reached and then calculate the area of the right triangle formed.
Student working
h² + 5² = 13²✓
h² = 169 − 25 = 144; h = 12 m✓
Area = 5 × 12 = 60 m²~
Student’s final answer
12 m; 60 m²
Marking breakdown
1/1
The ladder is correctly identified as the hypotenuse.
2/2
Correct subtraction and positive square root.
1/2
The base and perpendicular height are correct, but a triangle needs the factor ½.
Correct answer
12 m; 30 m²
Step-by-step explanation +
Pythagoras gives h = √144 = 12 m. The triangle’s area is ½ × 5 × 12 = 30 m²; 60 m² would be the corresponding rectangle’s area.
Solve the simultaneous equations 2x + y = 11 and x − y = 1.
Student working
Add the equations: 3x = 12✓
x = 4✓
4 − y = 1, so y = 3✓
Check: 2(4) + 3 = 11; 4 − 3 = 1✓
Student’s final answer
x = 4, y = 3
Marking breakdown
2/2
Adding eliminates y and gives the correct equation in x.
1/1
Correct first variable.
1/1
Correct substitution and second variable.
1/1
Both original equations are satisfied.
Correct answer
x = 4, y = 3
Step-by-step explanation +
The y coefficients are already opposites, so addition is efficient. Check both equations because satisfying only one does not establish a simultaneous solution.
In a right-angled triangle, the sides opposite and adjacent to angle θ are 6 cm and 8 cm. Find θ to one decimal place and the triangle’s hypotenuse.
Student working
tan θ = 6/8 = 0.75✓
θ = tan(0.75) = 0.0° to 1 d.p.×
Hypotenuse = √(6² + 8²)✓
√100 = 100 cm×
Student’s final answer
0.0°; 100 cm
Marking breakdown
2/2
Correct trigonometric ratio and substitution.
0/1
Use inverse tangent to recover the angle, not tangent.
1/1
Correct use of Pythagoras.
0/1
The square root of 100 is 10.
Correct answer
36.9°; 10 cm
Step-by-step explanation +
In degree mode, θ = tan⁻¹(0.75) = 36.869…°, which rounds to 36.9°. The hypotenuse is √100 = 10 cm. It should be longer than either leg but not longer than their sum.
Find the equation of the line through A(2, 5) and B(6, 13). Find the midpoint of AB.
Student working
m = (13 − 5)/(6 − 2) = 2✓
5 = 2(2) + c, so c = 1✓
y = 2x + 1✓
Midpoint = ((2 + 6)/2, (5 + 13)/2) = (4, 9)✓
Student’s final answer
y = 2x + 1; midpoint (4, 9)
Marking breakdown
2/2
Correct gradient method and value.
1/1
Correct intercept from substitution.
1/1
Correct line equation.
1/1
Both coordinates are averaged correctly.
Correct answer
y = 2x + 1; midpoint (4, 9)
Step-by-step explanation +
Gradient is change in y divided by change in x. Substitute either point to determine the intercept. The midpoint averages the horizontal and vertical coordinates separately.
A histogram has classes 0 ≤ t < 10 and 10 ≤ t < 30 with frequencies 15 and 20. Find their frequency densities. A third bar covers 30 ≤ t < 45 and has density 2.4. Find its frequency and the total frequency.
Student working
First density = 15/10 = 1.5✓
Second density = 20/20 = 1✓
Third frequency = (45 − 30) × 2.4 = 36✓
Total = 15 + 20 + 36 = 61×
Student’s final answer
1.5, 1; 36; total 61
Marking breakdown
1/1
Correct first class width.
1/1
Correct treatment of the wider class.
2/2
Correct width and density-to-frequency calculation.
0/1
The sum is 71.
Correct answer
1.5, 1; 36; total 71
Step-by-step explanation +
The histogram method is secure: frequency is bar area. Only the total is wrong. Grouping 15 + 20 = 35 before adding 36 gives 71.
Two similar solids have corresponding lengths in the ratio 2:3. The smaller has volume 40 cm³ and surface area 24 cm². Find the larger volume and surface area.
Student working
Length scale factor = 3/2✓
Volume = 40 × (3/2)³ = 135 cm³✓
Area = 24 × 3/2 = 36 cm²×
Student’s final answer
135 cm³; 36 cm²
Marking breakdown
1/1
The enlargement direction is correct.
2/2
Volume scales with the cube of the length factor.
0/2
Area requires the square of the length factor, giving 54 cm².
Correct answer
135 cm³; 54 cm²
Step-by-step explanation +
Use k for lengths, k² for areas and k³ for volumes. Here 24 × (3/2)² = 24 × 9/4 = 54 cm².
A rectangle measures 8 cm by 5 cm, each rounded to the nearest centimetre. State the error interval for each length and find the lower and upper bounds of its area.
Student working
7.5 ≤ l < 8.5✓
4.5 ≤ w < 5.5✓
Lower area = 7.5 × 4.5 = 33.75 cm²✓
Upper area = 8.5 × 4.5 = 38.25 cm²×
Student’s final answer
33.75 ≤ A < 38.25 cm²
Marking breakdown
1/1
Correct interval for the first length.
1/1
Correct interval for the second length.
2/2
Both lower bounds are used correctly.
0/2
The upper area needs both upper bounds: 8.5 × 5.5.
Correct answer
33.75 cm² ≤ A < 46.75 cm²
Step-by-step explanation +
With two positive lengths, increasing either length increases area. Use 8.5 × 5.5 = 46.75 cm² for the upper bound. The upper endpoint is excluded because both rounded lengths have open upper endpoints.
The difference-of-squares method earns one mark. The final subtraction should give 2.
Correct answer
8√2; 3√5/5; 2
Step-by-step explanation +
Part (c) is a difference of squares: (√3)² − 1² = 3 − 1 = 2. Keep exact forms throughout. The marks here follow the demonstration rubric shown, not an official board scheme.
For y = x² − 6x + 11, find dy/dx, find the stationary point, and state whether it is a minimum or maximum with a reason.
Student working
dy/dx = x − 6×
x − 6 = 0, so x = 6; y = 11~
Minimum because the x² coefficient is positive✓
Student’s final answer
dy/dx = x − 6; minimum at (6, 11)
Marking breakdown
0/2
The derivative of x² is 2x, not x.
1/3
Credit for setting the derivative to zero. The incorrect derivative gives the wrong stationary point.
1/1
The classification and reason are correct independently of the wrong coordinates.
Correct answer
dy/dx = 2x − 6; minimum at (3, 2)
Step-by-step explanation +
Apply the power rule to obtain 2x − 6. Setting it to zero gives x = 3. Substitute into the original function: 9 − 18 + 11 = 2. A positive x² coefficient means the parabola opens upwards.
Solve 2x² + 3x − 1 = 0. Give exact roots and decimal values correct to three decimal places.
Student working
a = 2, b = 3, c = −1✓
b² − 4ac = 9 − 4(2)(−1) = 17✓
x = (−3 ± √17)/4✓
x = 0.281 or x = −1.780~
Student’s final answer
(−3 ± √17)/4; 0.281, −1.780
Marking breakdown
1/1
All three coefficients have the correct signs.
2/2
Correct discriminant method and value.
2/2
Correct substitution, including the full denominator 2a.
1/2
The positive root is correctly rounded. The negative root rounds to −1.781.
Correct answer
(−3 ± √17)/4; 0.281, −1.781
Step-by-step explanation +
The exact roots are correct. √17 ≈ 4.123105626, so the roots are approximately 0.280776406 and −1.780776406. Round the final values, not the intermediate square root.
Of 30 students, 18 study French, 14 study Spanish and 8 study both. Find the number studying only French, only Spanish and neither. Find the probability of studying at least one language, and the probability of studying Spanish given that the student studies French.
Student working
French only = 18 − 8 = 10; Spanish only = 14 − 8 = 6✓
At least one = 18 + 14 − 8 = 24; neither = 30 − 24 = 6✓
P(at least one) = 24/30 = 4/5✓
P(Spanish given French) = 8/30 = 4/15×
Student’s final answer
10; 6; 6; 4/5; 4/15
Marking breakdown
2/2
The intersection is subtracted from each set total.
2/2
The intersection is counted once and the outside region is correct.
1/1
Correct probability from the whole group.
0/2
The condition restricts the denominator to the 18 students studying French.
Correct answer
10; 6; 6; 4/5; 4/9
Step-by-step explanation +
For the conditional probability, look only at the French group. Eight of its eighteen students also study Spanish, so P(Spanish | French) = 8/18 = 4/9.